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一个关于定义域光滑不变量的问题

Nekomusume
Nekomusume

This person is lazy, nothing was left behind...

我的提问：定理 22.3（定义域的光滑不变量）令$U \subset\mathbb{R}^n$为一个开子集，$S \subset\mathbb{R}^n$为一个任意子集，并且$f : U \rightarrow S$是一个微分同胚。那么$S$在$\mathbb{R}^n$中是开集。我无法理解为何集合$S$在$\mathbb{R}^n$中并不是自动开的。映射$f$是一个微分同胚，这意味着它在两个方向都是连续的，所以$S$是开的。回答：首先你所知道的是$U$中的开集$V$满足：$f(V)$在$S$中开，不是$f(V)$在$\mathbb{R}^n$中开。所以$f(U)=S$是在$S$中开。那个推断是说接着$f(U)=S$自动在$\mathbb R^n$中开，这是不一样的并且不是自动的。它需要证明。PS：这里说的是拓扑学中关于开集的一个重要盲点，即开集是相对的。尤其是考虑某个拓扑空间中的子集，要弄清楚究竟是在子集内开，还是在全空间内开。

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2024-11-06 19:32:51
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RicciflowsRicciflows
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An introduction to different branches of mathematics

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The note is mainly a sketch of basic knowledge concerning general topology, differential geometry, functional analysis, algebraic geometry, etc., starting from a discussion of Euclidean spaces. However, there maybe some mistakes in the note, so use at your own risk. For simplicity, some details are omitted and can be found in the references provided. Further materials concerning algebraic geometry, especially arithmetic algebraic geometry, can be referred to another note written by the author, N ...